Thread: Simple matrix problem

Here's what I've got so far. The augmented matrix for the system is:

3 1 1 1 b1
1 -1 1 1 b2
1 1 -1 1 b3
1 1 1 -1 b4

reduced eschelon form I made to be:

1 1/3 1/3 1/3 (1/3b1)
0 1 -1/2 -1/2 (-3/4b2 + 1/4b1)
0 0 1 -1 -b3 (-1/2b2 + 1/2b1)
0 0 0 0 (-5/6b1 + 3/4b2 + 4/3b3 + 1/2b4)

Which means -5/6b1 + 3/4b2 + 4/3b3 + 1/2b4 has to equal 0 for the system to be consistent, and the final answer should be:

[-5/6,3/4,4/3,1/2]

Is there a quick way to check the answer? Also, how do I solve the system with the values previously found, in the form:

|x|..... | b1 |
|y|..... | b2 |
|z| = C| b3 | + t * d
|v|......| b4 |

Thanks!

Well, now I now that this reduced echelon form

1 1/3 1/3 1/3 (1/3b1)
0 1 -1/2 -1/2 (-3/4b2 + 1/4b1)
0 0 1 -1 -b3 (-1/2b2 + 1/2b1)
0 0 0 0 (-5/6b1 + 3/4b2 + 4/3b3 + 1/2b4)

is wrong, but I don't know where I did something wrong... any help?